Forms of the axiom of choice

[mimimi]

Hello! During the winter break I started to study for myself the set theory and especially the axiom of choice. I found the next problem in a book:
Prove that the 3 statements of the axiom of choice are equivalent :

1) For any non-empty collection X of pairwise disjoint non-empty sets, there exists a choice set. pay attention, the book provides the next statement for the choice set the next statement:
X - non-empty collection of pairwise disjoint non-empty sets. Then Y is a choice set for X if Y is included in the reunion of all the elements of X and for any A, A included in X, there is a set H so that A ∩ Y = {H}. I suppose this is resembles more like what Halmos claimed. What I am trying to do is to prove all these implication using a formal language, with logical notations .

2) For any non-empty collection of non-empty sets X there is a choice function.

3) For any non-empty set X, there exists a function f(X)\{∅}→X so that for any non-empty set A⊆X, f(A) ∈ A.

Now, I have already tried and also succeeded to prove 2 of the 6 possible implications between the statements. But I simply can't realize how to prove the next implications: 1=>2, 2=>3 and 3=>1 using only the formal language, with logical notations and the Zermelo-Fraenkel axioms.

 

[pka]

Prove that the 3 statements of the axiom of choice are equivalent :
1) For any non-empty collection X of pairwise disjoint non-empty sets, there exists a choice set. pay attention, the book provides the next statement for the choice set the next statement:
X - non-empty collection of pairwise disjoint non-empty sets. Then Y is a choice set for X if Y is included in the reunion of all the elements of X and for any A, A included in X, there is a set H so that A ∩ Y = {H}. I suppose this is resembles more like what Halmos claimed. What I am trying to do is to prove all these implication using a formal language, with logical notations .

2) For any non-empty collection of non-empty sets X there is a choice function.

3) For any non-empty set X, there exists a function f(X)\{∅}→X so that for any non-empty set A⊆X, f(A) ∈ A.

Zermelo's axiom of choice: given any set of mutually disjoint nonempty sets, there exists at least one set that contains exactly one element in common with each of the nonempty sets.
I suppose that is a choice set.
So 1 => 2. Say Y is the choice set in #1, If A is in the collection define $\displaystyle f(A)=x, :x\in A\cap Y$ then $\displaystyle f(A)\in A.$

2=> 3. If $\displaystyle \mathcal{C}$ is a choice function and $\displaystyle A\in P(X)\setminus\{\emptyset\}$ then define $\displaystyle f(A)=\mathcal{C}(A)\in A$.

3=>1 Using the function $\displaystyle f$ given in #3 $\displaystyle \forall A\in \mathcal{X}$ define $\displaystyle Y=\{x:x\in f(A)\}$.